Educational Codeforces Round 26:C. Two Seals

题目：

One very important person has a piece of paper in the form of a rectanglea × b.

Also, he has n seals. Each seal leaves an impression on the paper in the form of a rectangle of the sizex_i × y_i. Each impression must be parallel to the sides of the piece of paper (but seal can be rotated by 90 degrees).

A very important person wants to choose two different seals and put them two impressions. Each of the selected seals puts exactly one impression. Impressions should not overlap (but they can touch sides), and the total area occupied by them should be the largest possible. What is the largest area that can be occupied by two seals?

Input

The first line contains three integer numbers n, a and b (1 ≤ n, a, b ≤ 100).

Each of the next n lines contain two numbersx_i,y_i (1 ≤ x_i, y_i ≤ 100).

Output

Print the largest total area that can be occupied by two seals. If you can not select two seals, print0.

Examples

Input

2 2 21 22 1

Output

4

Input

4 10 92 31 15 109 11

Output

56

Input

3 10 106 67 720 5

Output

0

Note

In the first example you can rotate the second seal by 90 degrees. Then put impression of it right under the impression of the first seal. This will occupy all the piece of paper.

In the second example you can't choose the last seal because it doesn't fit. By choosing the first and the third seals you occupy the largest area.

In the third example there is no such pair of seals that they both can fit on a piece of paper.

题意：给出n张纸片的长和宽，再给出一个矩形的长和宽（a和b），求两张图片在矩形上放置的最大面积

思路：有两种情况 ，将第一张纸片横着放 or 竖着放，剩下的可以划分成3小块（再把其中相连的2块合成1块），判断第二张纸片能否不重叠地在放在矩形上。自己动手画个图就知道了。数据不打，暴力可过~

CODE:

#include<bits/stdc++.h>using namespace std;struct node{        int x,y;}q[105];int n,a,b;int fun(){        int i,j,x1,x2,x3,x4,y1,y2,y3,y4,maxr=0,flag;        for(i=0;i<n;i++){                x1=q[i].x;y1=q[i].y;   //纸片一                for(j=i+1;j<n;j++){                        flag=0;   //判断是否已选择一种方式存放                        x2=q[j].x;y2=q[j].y;   //纸片二                        if(x1<=a&&y1<=b){                                x3=a-x1;y3=b;                                x4=b-y1;y4=a;                                if(x2<=x3&&y2<=y3||x2<=x4&&y2<=y4||x2<=y3&&y2<=x3||x2<=y4&&y2<=x4){                                        maxr=max(maxr,x1*y1+x2*y2);                                        flag=1;                                }                        }                        if(!flag){                                if(x1<=b&&y1<=a){                                        x3=a-y1;y3=b;                                        x4=b-x1;y4=a;                                        if(x2<=x3&&y2<=y3||x2<=x4&&y2<=y4||x2<=y3&&y2<=x3||x2<=y4&&y2<=x4) maxr=max(maxr,x1*y1+x2*y2);                                }                        }                }        }        return maxr;}int main(){        while(~scanf("%d%d%d",&n,&a,&b)){                if(a>b) swap(a,b);                for(int i=0;i<n;i++){                        scanf("%d%d",&q[i].x,&q[i].y);                        if(q[i].x>q[i].y) swap(q[i].x,q[i].y);                }                printf("%d\n",fun());        }        return 0;}